Answer:
The following query is used to display TripName. Reservationld, FirstName. LastName and TotalCost of trip by adding the trip price plus other fees and multiplying the result by the number of persons which have number of persons >4.
Query:
SELECT ReservationlD, Trip.TripName. Customer.LastName. Customer.FirstName. (TripPrice+OtherFees) 'NumPersons as TotalCost FROM Reservation, Trip, Customer WHERE NumPersons>4 AND Reservation.TriplD=Trip.TriplD AND
Customer. CustomerNum=Reservation.CustomerNum:
Explanation:
- Select clause is used to retrieve data from specified database table or relation and returns the data in the form of table.
- ReservationID. Trip.TripName. Customer.LastName. Customer.FirstName are the column name of table.
- (TripPrice+OtherFees) 'NumPersons will calculate the total cost of the Trip and stored it into TotalCost column.
- As clause is used to give new name TotalCost to resultant column.
- FROM clause specifies one or more table from where records to be retrieved. o Reservation. Trip and Customer are the table name.
- WHERE clause is used in SQL query to retrieve only those records that satisfy the specified condition
The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
<h3>
Thickness of the aluminum</h3>
The thickness of the aluminum can be determined using from distance of closest approach of the particle.

where;
- Z is the atomic number of aluminium = 13
- e is charge
- r is distance of closest approach = thickness of aluminium
- k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

<h3>For 2.5 MeV protons</h3>
Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.
<h3>For 10 MeV alpha-particles</h3>
Charge of alpah particle = 2e

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.
Learn more about closest distance of approach here: brainly.com/question/6426420
Answer:
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Explanation:
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Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
Its 0.001
0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m